Question

Hydrogen cyanide is used to prepare sodium cyanide, which is used in part to obtain gold from gold-containing rock. If a reac- tion vessel contains 11.5 g NH3, 12.0 g O2, and 10.5 g CH4, what is the maximum mass in grams of hydrogen cyanide that could be made, assuming the reaction goes to completion as written?

Answer 1

Answer:

Mass of HCN produced = 6.75 g

Explanation:

Reaction is as follows:

$2NH_3+3O_2+2CH_4 \rightarrow 2HCN + 6H_2O$

First calculate the no. of moles of each chemical species.

molecular mass of $NH_3$ is 17 g/mol

No. of mol of $NH_3$ = 11.5/17 = 0.676 mol

Molecular mass of $O_2$ = 32 g/mol

No. of  mol of $O_2$ = 12/32 = 0.375

Molecular mass of $CH_4$ = 16 g/mol

No. of  mol of $CH_4$ = 10.5/16 = 0.656 mol

from the balanced chemical reaction, it is clear that 2-moles ammonia reacts with 3 moles oxygen and 2 moles methane to form 2 moles of HCN.

or, 1-mol ammonia reacts with 1.5 mol oxygen and 1 mol methane to form 1 mol of HCN.

Thus, no. of oxygen present is less than required and so it will act as limiting reagent.

From the chemical equation,

3 moles oxygen produces 2 moles HCN

or one mole oxygen produces (2/3) moles HCN

0.375 moles oxygen produces (2/3) × 0.375 HCN = 0.25 moles of HCN  

molar mass of HCN = 27 g/mole

Mass = mol × molar mass

mass of HCN = 27 × 0.25 = 6.75 g